44 111 METRIC SPACES

Indeed, if (Gn) is an at most denumerable basis for the open sets of E,
the sets Gn n F form a denumerable basis for the open sets of F <= E, due
to (3.10.1) and (1.8.2). Hence the result by (3.9.4).

(3.10.10) Let A be a subset of a metric space E; a point x0 e A is said to be
isolated w. A if there is a neighborhood V of x0 in E such that V n A = {x0}.
To say that every point of A is isolated in A means that in the subspace A
every set is an open set (in other words the subspace A is homeomorphic to a
discrete space: see Section 3.12). In a separable metric space, every subset
all of whose points are isolated is therefore at most denumerable (3.9.4).

PROBLEMS

1. Let B, B' be two nonempty subsets of a metric space E, and A a subset of B n B',
which is open (resp. closed) both with respect to B and with respect to B'; show that A
is open (resp. closed) with respect to B u B',

2. Let (Ua) be a covering of a metric space E, consisting of open subsets. In order that a
subset A of E be closed in E, it is necessary and sufficient that each set A n Ua be closed
with respect to U*.

3. In a metric space E, a subset A is said to be locally closed if for every x e A, there is a
neighborhood V of x such that A n V is closed with respect to V. Show that the locally
closed subsets of E are the sets U n F, where U is open and F closed in E. (To prove
that a locally closed set has that form, use Problem 2.)

4. Give an example of a subspace E of the plane R2, such that there is in E an open ball
which is a closed set but not a closed ball, and a closed ball which is an open set but not
an open ball. (Take E consisting of the two points (0,1) and (0, — 1) and of a suitable
subset of the joaxis.)

5. Give a proof of (3.10.9) without using the notion of basis (in other words, exhibit
an at most denumerable subset which is dense in the subspace).

11. CONTINUOUS MAPPINGS

Let E and E' be two metric spaces, d, d' the distances on E and E'. A
mapping/of E into E' is said to be continuous at a point x0 e E if, for every
neighborhood V of/(#<>) in E', there is a neighborhood V of x0 in E such
that/(V) cr V';/is said to be continuous in E (or simply "continuous") if it
is continuous at every point of E.

If we agree that the mathematical notion of neighborhood corresponds to
the intuitive idea of" proximity," then we can express the preceding definition
in a more intuitive way, by saying that f(x) is arbitrarily close to f(xQ) as
soon as x is close enough to x0.umerable open covering.nct.